3 unknowns, 3 equations, failure on algebraic method

In summary: I meant to say "If you still can't work it out, you need to post details of your work, so we can help you."
  • #1
togo
106
0

Homework Statement



This example comes straight out of the book. Can someone show intermediate steps?

Code:
Fx = 0: -0.0933Tab - 0.0846Tac + P = 0
Fy = 0: +0.778Tab + 0.705Tac - 1962N = 0
Fz = 0: +0.622Tab - 0.705Tac = 0

Book solution: P = 235 N, Tab = 1401N, Tac = 1236N

Homework Equations



what method did they use

The Attempt at a Solution



determinants & algebraic methods failed.
 
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  • #2
togo said:

Homework Statement



This example comes straight out of the book. Can someone show intermediate steps?

Code:
Fx = 0: -0.0933Tab - 0.0846Tac + P = 0
Fy = 0: +0.778Tab + 0.705Tac - 1962N = 0
Fz = 0: +0.622Tab - 0.705Tac = 0

Book solution: P = 235 N, Tab = 1401N, Tac = 1236N

Homework Equations



what method did they use

The Attempt at a Solution



determinants & algebraic methods failed.
I would suggest you check your hand calculation or calculator work more carefully. Those values does work in the equations.

Try adding the last two equations together: you should get 1.4 Tab = 1962 N , which does give the book's answer. Now 0.622 Tab = 0.705 Tac , so Tac = (0.622/0.705) Tab . Putting these values into the first equation will let you find P .

It would help in troubleshooting if you showed what you'd done...
 
  • #3
togo said:

Homework Statement



This example comes straight out of the book. Can someone show intermediate steps?

Code:
Fx = 0: -0.0933Tab - 0.0846Tac + P = 0
Fy = 0: +0.778Tab + 0.705Tac - 1962N = 0
Fz = 0: +0.622Tab - 0.705Tac = 0

Book solution: P = 235 N, Tab = 1401N, Tac = 1236N

Homework Equations



what method did they use

The Attempt at a Solution



determinants & algebraic methods failed.

The problem is sensitive to roundoff errors. When I solve the system in Maple 14 I get Tab = 1401.42857142857, Tac = 1236.43768996960 and P = 235.355914285714. Putting these in a column vector X and computing A.X gives [0, 1962, .1137e-12] (taking N=1) which is close to the actual right-hand-sides [0,1962,0]. However, rounding off to Tab = 1401, Tac = 1236, P=235 in a vector X gives A.X = [-.2789, 1961.358, .0420], which is not particularly close to [0,1962,0]. I don't know if this is what you are getting, as you have not explained exactly what you mean by failure.

RGV
 
  • #4
Ok I posted the answer to these equations in my original post, but I wasn't trying to verify whether or not they work. They do work; what I posted is the last part of the example in my book. What I want to know is how they took on the 2nd last step those equations and somehow got that result?

There appears to be 3 unknowns: Tab, Tac, and P. In another thread on another question I tried using Cramers rule to solve a similar question using determinants. Next I tried the algebraic solution to try and arrive at the book's result, but with 3 unknowns I kept going in circles. I ended up with 2 unknowns no matter what I did.

The book just says ''solving these equations we obtain'' and then lists the result. Can someone please explain how to solve these equations? thanks.
 
Last edited:
  • #5
togo - I would start with the last two equations. Note that they contain just two variables (Tab and Tac).

Use the last equation and solve for Tab in terms of Tac (or Tac in terms of Tab).


Next (I'll assume you solved for Tab in terms of Tac):

Substitute Tab expresion into the second equation so you get an equation soley in terms of Tac. Solve this equation for Tac.

Now, using the expresion you first solved for (Tab as a function of Tac), use the value of Tac and sove for the value of Tab.

Use the values for Tab and Tac in the first equation to solve for the value of P.
 
  • #6
togo said:
The book just says ''solving these equations we obtain'' and then lists the result. Can someone please explain how to solve these equations? thanks.
dynamicsolo already told you how to solve the equations step by step.

If you still can't work it out on your own, you need to post details of your work, so we can see where you're going wrong.
 
  • #7
crap your right, sorry.
 

FAQ: 3 unknowns, 3 equations, failure on algebraic method

What does the term "3 unknowns, 3 equations" mean?

The term "3 unknowns, 3 equations" refers to a system of equations that contains three unknown variables and three equations that can be used to solve for those variables. This type of system is also known as a "three-by-three" system.

What is the algebraic method used to solve a system of equations?

The algebraic method, also known as the substitution method, involves solving one of the equations for one of the unknown variables and then substituting that expression into the other equations to eliminate that variable. This process is repeated until all unknown variables have been solved for.

Why might the algebraic method fail to solve a system of equations?

The algebraic method may fail to solve a system of equations if the equations are not independent, meaning that one or more of the equations can be derived from the others. In this case, there are not enough equations to uniquely determine the values of all unknown variables.

Are there alternative methods to solve a system of equations?

Yes, there are alternative methods to solve a system of equations, such as the graphing method, elimination method, and matrices method. These methods may be more suitable for certain types of systems or may be more efficient than the algebraic method.

How can I determine if a system of equations has a unique solution?

A system of equations will have a unique solution if the number of unknown variables is equal to the number of equations and the equations are independent. This means that there is exactly one solution that satisfies all of the equations in the system.

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